TITLE: Let $G$ be a finite abelian group, and let $n$ divide $|G|$. Let $m$ be the number of solutions of $x^n=1$. Prove that $n\mid m$.
QUESTION [6 upvotes]: Let $G$ be a finite abelian group, and let $n$ divide $|G|$.  Let $m$ be the number of solutions of $x^n=1$.  Prove that $n\mid m$.

My attempt
It's tempting to find a way to use Lagrange's theorem.  Maybe something here is a subgroup of something else?  We can fix $n$ and take the subgroup of $G$ of all elements which solve $x^n=1$.  Proof that this is a subgroup:  Inverses of solutions are always solutions.  Because the group is abelian, products of solutions are solutions.  QED.
Great, so it's a subgroup, so $m$ divides the order of $G$.  So does $n$.  I'm not sure that this really got me anywhere.  It'd be nice if there were some relevant subgroup of order $n$.
Being finite and abelian then it has a representation as $G\cong C_{p_1^{n_1}}\times\dots\times C_{p_k^{n_k}}$, a product of cyclic groups of prime power order.  The solutions are exactly the product of solutions "in each factor", i.e. solutions of the form $\langle e, \dots, e, x, e, \dots, e\rangle$ where $x\in C_{p_i^{k_i}}$ for some $i$.  So perhaps something comes from thinking about the number of solutions to $x^n=1$ where $x$ is taken from $C_{p_i^{k_i}}$.
Again this is a subgroup so the number of solutions divides $p_i^{k_i}$, and $p_i^{k_i}$ divides $|G|$. And $n$ divides the order of $G$.  But at this point I'm not sure whether I'm on a productive path, since these facts don't seem to be enough to show that $n|m$.
In fact the more that I think about how $n$ is so-to-speak missing factors from $|G|$ the more I think that finding numbers which divide $|G|$ just isn't a productive path.

REPLY [3 votes]: By the fundamental theorem of finite abelian groups we may choose a $G$ subgroup $G_n$ of size $n$. Lagrange's theorem gaurantees $G_n\leq\ker(\varphi_n)$ where $\varphi_n$ denotes the $G$ endomorphism $\varphi_n:x\mapsto x^n$ and $\leq$ denotes subgroup inclusion. Finally, by Lagrange's theorem once again, $$n=|G_n|\;\Big\vert\;|\ker(\varphi_n)|=m$$